d and f Block Elements

$X\buildrel \Delta \over \longrightarrow {O_2} + Y + Z$ and $Z\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{KN{O_3}}^{KOH}} Y$

$\mathop {2KMn{O_4}}\limits_{(X)} \buildrel {} \over \longrightarrow \mathop {{K_2}Mn{O_4}}\limits_{(Y)} + \mathop {Mn{O_2}}\limits_{(Z)} + {O_2}$

$\mathop {Mn{O_2}}\limits_{(Z)} \mathrel{\mathop{\kern0pt\longrightarrow} \limits_{KN{O_3}}^{KOH}} \mathop {{K_2}Mn{O_4}}\limits_{(Y)} $

$\therefore$ $X = KMn{O_4},Y = {K_2}Mn{O_4}$ and $Z = Mn{O_2}$

(A) Electronic configuration of $\mathrm{Cu}$ is

$1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^1$

$\Delta_i H_1$ is low while $\Delta_i H_2$ is highest because of fully filled $d$-orbital.

(B) Electronic configuration of $\mathrm{Zn}$ is $1 s^2 2 s^2 2 p^6 3 s^2 3 p^6 3 d^{10} 4 s^2$

First 2 electrons are removed from $s$-orbital while 3rd electron released from fully filled $d$-orbital, which require large amount of energy.

(C) $M$ in $\left[M(\mathrm{CO})_6\right]$ is $\mathrm{Cr}$.

(D) Ni has highest heat of atomisation in the $3 d$-transition series.

(a) Basic character of metal oxides decreases with increase in oxidation number

$\stackrel{+2}{M \mathrm{O}}, \stackrel{+3}{M_2}\mathrm{O}_3, \stackrel{+4}{M} \mathrm{O}_2, \stackrel{+5}{M_2} \mathrm{O}_5$
(Decreasing basics character)

(b) Manganese ($\mathrm{Mn}$) has maximum number of oxidation state.

Sc, V, Cr, Mn (Increasing number of oxidation states)

(c) Magnetic moment $=\sqrt{n(n+1)}$

( $n=$ number of unpaired electron)

Magnetic moment $\propto$ unpaired electron

So, the correct order of magnetic moment is

$d^5>d^4>d^3>d^1$

(d) $\begin{aligned} \mathrm{Mn}^{2+} & \rightarrow 3 d^5 \\ \mathrm{Fe}^{2+} & \rightarrow 3 d^6 \\ \mathrm{Cr}^{2+} & \rightarrow 3 d^4 \\ \mathrm{Co}^{2+} & \rightarrow 3 d^7 \end{aligned}$

$\mathrm{Mn}^{2+}$ is most stable as $d$-subshell is half filled.

The electronic configuration of $\mathrm{Cu}^{2+}$ is

$[\mathrm{Ar}] 3 d^9$

It contains one unpaired electron,

Hence, $\mathrm{CuCl}_2$ is coloured compound. In all other compounds,

$\begin{aligned} & \mathrm{CuCl} \rightarrow\left[\mathrm{Cu}^{+}\right] \Rightarrow 3 d^{10} 4 s^0 \\ & \mathrm{ScCl}_3 \rightarrow\left[\mathrm{Sc}^{3+}\right] \Rightarrow 3 p^6 4 s^2 \\ & \mathrm{TiCl}_4 \rightarrow\left[\mathrm{Ti}^{4+}\right] \Rightarrow 3 p^6 4 s^2 \end{aligned}$

No unpaired electrons are present.

For general electronic configuration of metal,

$\mathrm{Fe}=1 s^2, 2 s^2 2 p^6, 3 s^2 3 p^6 3 d^6, 4 s^2$

Now, for $\mathrm{Fe}=[\mathrm{Ar}] 3 d^6 4 s^2$

$\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 d^6 4 s^0$ [Remove $2 e^{-}$ from $s$-orbital first]

$\Rightarrow \quad \mathrm{Fe}^{3+}=[\mathrm{Ar}] 3 d^6 4 s^0$

$\mathrm{Fe}^{3+}$ is denoted as ferric (III) ion.

Fe(Z = 26) = [Ar]3d$^6$ 4s$^2$

Fe$^{2+}$ = [Ar]3d$^6$ 4s$^0$

Total number of unpaired electrons = 4

i.e., n = 4

Magnetic moment, $\mu=\sqrt{n(n+2)}$

$=\sqrt{4(4+2)}=\sqrt{4\times6}$

$=\sqrt{24}$

$=4.9$ BM

Hence, magnetic moment of Fe$^{2+}$ is 4.9 BM.

$\mathrm{CrO}_4^{2-}$ is a stronger oxidising agent as compared to $\mathrm{MnO}_4^{2-}$.

$\mathrm{Cr}^{+6} \rightarrow[\mathrm{Ar}] 3 d^0 4 s^0$ which is stable. So, Cr reading loses its 6 electrons to attain stable noble gas and acting as oxidising agent but in $\mathrm{Mn}^{+6}$ [electronic configuration is $[\mathrm{Ar}] 3 d^1 4 s^0$ ] one electron is difficult to remove and acquire stable form. Hence, $\mathrm{MnO}_4^{-}$ is not oxidising agent but $\mathrm{CrO}_4{ }^{2-}$ is good oxidising agent.

The given electronic configuration is [Ar] 3d$^5$ 4s$^2$. To determine the group of the periodic table to which this element belongs, we need to analyze its electron configuration, particularly the electrons in the outermost shells.

In this configuration:

• [Ar] represents the argon core, which is a stable, noble gas with 18 electrons.
• 3d$^5$ indicates there are 5 electrons in the 3d subshell.
• 4s$^2$ indicates there are 2 electrons in the 4s subshell.

Total electrons in the outer shells (valence electrons) = 5 (in 3d subshell) + 2 (in 4s subshell) = 7 valence electrons.

This pattern of electron configuration is characteristic of the elements in Group 7 of the periodic table. The 3d subshell is filling and the outer 4s subshell contains 2 electrons, typical of a transition metal in Group 7. Examples of such elements include manganese (Mn), which has the atomic number 25 and the electron configuration [Ar] 3d$^5$ 4s$^2$.

Therefore, the element with the electronic configuration [Ar] 3d$^5$ 4s$^2$ belongs to:

Option C: Seventh

To identify the incorrect statement, let's analyze each option :

Option A : "In manganate and permanganate ions, the π-bonding takes place by overlap of p-orbitals of oxygen and d-orbitals of manganese." This statement is true. In both manganate ${MnO}_4^{2-} $ and permanganate ${MnO}_4^- $ ions, the manganese atom is in a high oxidation state (+6 and +7, respectively) and forms bonds with oxygen atoms involving the overlap of p-orbitals of oxygen and d-orbitals of manganese.

Option B : "Manganate ion is green in colour and permanganate ion in purple in colour." This statement is true. Manganate ions are indeed green and permanganate ions are purple in aqueous solution.

Option C : "Manganate and permanganate ions are paramagnetic." This statement is incorrect. Both manganate and permanganate ions are diamagnetic, not paramagnetic. In these ions, manganese is in +6 and +7 oxidation states, respectively, and has no unpaired electrons.

Option D : "Manganate and permanganate ions are tetrahedral." This statement is true. Both manganate and permanganate ions have a tetrahedral geometry, with the manganese atom at the center and the four oxygen atoms surrounding it.

Therefore, the incorrect statement is option C.

$\mathrm{La}^{3+}(Z=57)$ has 54 electrons.

$\mathrm{Ti}^{3+}(Z=22)$ has 19 electrons.

$\mathrm{Lu}^{3+}(Z=71)$ has 68 electrons.

and $\mathrm{Sc}^{3+}(Z=21)$ has 18 electrons.

The species with unpaired electrons show colour in aqueous solution.

Because, $\quad \mathrm{Ti}^{3+}=[\mathrm{Ar}] \cdot 4 s^1$

Thus, has one unpaired electron.

Hence, option (b) is the correct answer.

Trans uranium elements are elements after uranium, i.e. trans uranium elements have atomic number, $Z>92$.

The highest oxidation state observed in the first row transition metals is +7 .

In first row ' $\mathrm{Mn}^{\prime}$ with outer configuration : $3 d^5, 4 s^2$ show +7 oxidation state, i.e.

It has 7 electrons in outermost orbital, so has +7 oxidation state.

Among the given series of transition metal ions, one in which all metal ions have $3 d^2, 3 p^6$ electronic configuration is :

Option (d), i.e.

$\mathrm{Ti}(Z=22) \rightarrow \mathrm{Ti}^{2+}$ (has 20 electrons)

$\mathrm{V}(\mathrm{Z}=23) \rightarrow \mathrm{V}^{3+}$ (has 20 electrons)

$\mathrm{Cr}(Z=24) \rightarrow \mathrm{Cr}^{4+}$ (has 20 electrons)

and $\mathrm{Mn}(Z=25) \rightarrow \mathrm{Mn}^{5+}$ (has 20 electrons)

Thus, all can show $3 d^2 3 p^6$ electronic configuration i.e. option (d) is the correct answer.

Interstitial compounds are formed, when small atoms are trapped inside the crystal lattice of metals.

(i) They are very hard and rigid.

(ii) They have high melting point, which are higher than these of the pure metals.

(iii) They show conductivity like the of the pure metals.

(iv) They aquire chemical inertness.

The highest possible oxidation states of uranium and plutonium can be identified by considering the electronic configurations of these elements and noting the number of valence electrons available for bonding.

Uranium (U, atomic number 92) has the electronic configuration: $ [Rn] 5f^3 6d^1 7s^2 $

Plutonium (Pu, atomic number 94) has the electronic configuration: $ [Rn] 5f^6 7s^2 $

To find the highest oxidation state, we sum the number of $d$, $f$, and $s$ electrons in the valence shell.

For uranium :

$ 3 (f) + 1 (d) + 2 (s) = 6 $

So the highest oxidation state of uranium is +6.

For plutonium :

$ 6 (f) + 2 (s) = 8 $

So the highest oxidation state of plutonium is +8. However, in practice, plutonium is known to exhibit a highest oxidation state of +7.

Comparing with the given options, the correct answer is :

Option D : 6 and 7.

Cu²⁺ and Ni²⁺ ions belong to Group II and Group IV respectively.

Group II sulphides precipitated out even under low concentration of sulphides, while Group IV sulphides require higher concentration of sulphides. To fulfill this reaction condition, dil. HCl is chosen in Group II reagents and NH₄OH is chosen in Group IV reagents.

Statement wise explanation is

Statement (a) Mn²⁺ produces yellow-green colour in flame test while Cu²⁺ produces bluish-green colour in flame test. Thus, due to the presence of green colour in both the cases, flame test is not the suitable method to distinguish between nitrate salts of Cu²⁺ and Mn²⁺. Hence this statement is wrong.

Statement (b) Cu²⁺ belong to group II of cationic or basic radicals. It gives black ppt. of CuS if H₂S is passed through it in the presence of acid (e.g. HCl). Mn₂₊ does not show this property hence this can be considered as a suitable method to distinguish between Mn²⁺ and Cu²⁺. Hence, this statement is correct.

Statement (c) In faintly basic medium when H₂S is passed both Cu²⁺ and Mn²⁺ forms precipitates. Thus, it is not suitable method to distinguish between them. Hence, this statement is incorrect.

Statement (d) The standard reduction potential of Cu²⁺ /Cu is +0.34 V while that of Mn²⁺/Mn is $-$1.18 V. This can be used to distinguish between Cu²⁺ and Mn²⁺. In general less electropositive metals have higher SRP. Hence, this statement is correct.